In a $T_0$ space the union of two scattered sets is scattered A scattered space if a space $X$ that contains no nonempty dense-in-itself subset.  Equivalently, every nonempty subset $A$ of $X$ contains a point isolated in $A$.
Note that in general the union of two scattered sets is not scattered.  For example, if $X=\{a,b\}$ with the indiscrete topology, $\{a\}$ and $\{b\}$ are both scattered, but their union, $X$, is not scattered as it has no isolated point.
It is shown in Kuratowski's book (p. 79) that in a $T_1$ space, the union of two scattered sets is scattered (he assumes spaces are $T_1$ unless mentioned otherwise).  I think the following is more generally true:

Theorem: In a $T_0$ space, the union of two scattered sets is scattered.

How can this be proved?
 A: Suppose $X$ is $T_0$.  Let $A$ and $B$ be two nonempty scattered subsets of $X$ and assume by contradiction that $A\cup B$ is not scattered.  So we can find a nonempty $C\subseteq A\cup B$ with $C$ dense-in-itself.
$C$ cannot be entirely contained in $B$, otherwise it would have an isolated point.  So $C\setminus B$ is a nonempty subset of $A$ (scattered) and has an isolated point $a$.  There is an open nbhd $V$ of $a$ such that $V\cap(C\setminus B)=\{a\}$.
But $a$ cannot be an isolated point of $C$.  So $V$ meets $C\cap B$, which must be a nonempty subset of $B$.  Since $B$ is scattered, $C\cap B$ has an isolated point $b$.  So we can find an open $U\subseteq V$ such that $U\cap(C\cap B)=\{b\}$.

*

*Case $a\notin U$.  Then $U\cap C=\{b\}$ and $C$ has an isolated point.  Contradiction.

*Case $a\in U$. Then $U\cap C=\{a,b\}$.  Since $X$ is $T_0$, there is an open set containing one of the points and not the other.  Intersecting further with $U$ shows that one of the two points is isolated in $C$.  Again, contradiction.

A: This, and a number of similar results, are proven in Scattered spaces, compactifications and an application to image classification problem by M. Al-Hajri et. al.
The proof they give is roughly as follows:
Let $A$ and $B$ be scattered subspaces, and $S\subseteq A\cup B$. We show that $S$ has an isolated point. Let $S_A \ = S\cap A$ and $S_B = S\cap B$, and note that $S_A$ and $S_B$ are scattered. Since $S_A$ is scattered, there is an $a\in S_A$ and and an open set $U$ such that $\{a\} = S_A\cap U$. If $U\cap S_B = \emptyset$, we're done.
So assume not. Since $U\cap S_B \subseteq S_B$, it has an isolated point. So there exists $b$ in $U\cap S_B$ and an open set $V$ such that $\{v\} = U\cap S_B \cap V$.
Thus the open set $U\cap V$ is either $\{b\}$ (in which case we're done), or $U\cap V = \{a,b\}$, in which case we use the fact that our space is $T_0$ to isolate one of $a$ or $b$, completing the proof.
